What is the integral of 1/(1+e^{ax)} ?

The integral of 1/(1+e^{ax)} is 1/a (-ln|1+e^{ax}|+ax)+C

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integral of 1/(1+e^{ax)}

\int \frac{1}{1 + e ^{a x}} d x

\frac{1}{a} (- ln ∣ 1 + e^{a x} ∣ + a x) + C

Solution steps

\int \frac{1}{1 + e ^{a x}} d x

Apply u-substitution

: \int \frac{1}{a u ( u - 1 )} d u

= \int \frac{1}{a u ( u - 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{a} \cdot \int \frac{1}{u ( u - 1 )} d u

Take the partial fraction of

\frac{1}{u ( u - 1 )} : - \frac{1}{u} + \frac{1}{u - 1}

= \frac{1}{a} \cdot \int - \frac{1}{u} + \frac{1}{u - 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{a} (- \int \frac{1}{u} d u + \int \frac{1}{u - 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= \frac{1}{a} (- ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = 1 + e^{a x}

= \frac{1}{a} (- ln ∣ 1 + e^{a x} ∣ + ln ∣ 1 + e^{a x} - 1 ∣)

Simplify

\frac{1}{a} (- ln ∣ 1 + e^{a x} ∣ + ln ∣ 1 + e^{a x} - 1 ∣) : \frac{1}{a} (- ln ∣ 1 + e^{a x} ∣ + a x)

= \frac{1}{a} (- ln ∣ 1 + e^{a x} ∣ + a x)

Add a constant to the solution

= \frac{1}{a} (- ln ∣ 1 + e^{a x} ∣ + a x) + C

\int \frac{x}{x ^{4} + 16} d x

2 y^{^{'''}} - 3 y^{^{''}} - 18 y^{^{'}} - 8 y = 0

y = 3 x^{2} - 12, y = 0

\frac{d}{d x} (\frac{3 + sin ( x )}{3 x + cos ( x )})

y = e^{x}, y = e^{- 3 x}, x = - 2, x = 1

What is the integral of 1/(1+e^{ax)} ?
The integral of 1/(1+e^{ax)} is 1/a (-ln|1+e^{ax}|+ax)+C